Problem: Let $g(x)=\begin{cases} \dfrac{x-3}{\sqrt{x+13}-4}&\text{for }x\geq -13, x\neq 3 \\\\ k&\text{for }x=3 \end{cases}$ $g$ is continuous for all $x>-13$. What is the value of $k$ ? Choose 1 answer: Choose 1 answer: (Choice A) A $4$ (Choice B) B $0$ (Choice C) C $3$ (Choice D) D $8$
Solution: $\dfrac{x-3}{\sqrt{x+13}-4}$ is continuous for all $x>-13$ other than $x=3$, which means $g$ is continuous for all $x>-13$ other than $x=3$. In order for $g$ to also be continuous at $x=3$, the following equality must hold: $\lim_{x\to 3}g(x)=g(3)$ Since $g(3)=k$, we will obtain the above equality by letting $k=\lim_{x\to 3}g(x)$. So let's find $\lim_{x\to 3}g(x)$, come on! $\begin{aligned} &\phantom{=}\lim_{x\to 3}g(x) \\\\ &=\lim_{x\to 3}\dfrac{x-3}{\sqrt{x+13}-4} \gray{\text{This is the rule for }x\neq 3} \\\\ &=\lim_{x\to 3}\dfrac{x-3}{\sqrt{x+13}-4}\cdot\dfrac{\sqrt{x+13}+4}{\sqrt{x+13}+4} \gray{\text{Rationalize}} \\\\ &=\lim_{x\to 3}\dfrac{(x-3)(\sqrt{x+13}+4)}{x+13-4^2} \gray{\text{Simplify}} \\\\ &=\lim_{x\to 3}\dfrac{\cancel{(x-3)}(\sqrt{x+13}+4)}{\cancel{x-3}} \gray{\text{Cancel common factors}} \\\\ &=\lim_{x\to 3}(\sqrt{x+13}+4) \\\\ &\text{(This is allowed because }x\neq 3) \\\\ &=\sqrt{3+13}+4 \gray{\text{Direct substitution}} \\\\ &=8 \end{aligned}$ We obtained that if we set $k=8$, then $\lim_{x\to 3}g(x)=g(3)$, which makes $g$ continuous at $x=3$. Since we already saw that $g$ is continuous for any other $x>-13$, we can determine that it's continuous for all $x>-13$. In conclusion, $k=8$.